Let's start by computing the values of the product for small values of $n$.

The product in question is:

$P(n) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \cdots \left(1 + \frac{1}{n}\right)$

Step 1: Compute for small values of $n$.

• For $n = 1$: $P(1) = 1 + \frac{1}{1} = 2$

• For $n = 2$: $P(2) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) = 2 \times \frac{3}{2} = 3$

• For $n = 3$: $P(3) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) = 2 \times \frac{3}{2} \times \frac{4}{3} = 4$

• For $n = 4$: $P(4) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \left(1 + \frac{1}{4}\right) = 2 \times \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} = 5$

• For $n = 5$: $P(5) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \left(1 + \frac{1}{4}\right) \left(1 + \frac{1}{5}\right) = 2 \times \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \frac{6}{5} = 6$

From the pattern, we can conjecture that the general form of the product is:

$P(n) = n + 1$

Step 2: Prove by mathematical induction.

We will prove that $P(n) = n + 1$ by induction.

Base Case:

For $n = 1$: $P(1) = 2 = 1 + 1$ The base case holds true.

Inductive Step:

Assume that for some $k \geq 1$, the formula holds, i.e., $P(k) = k + 1$ We need to prove that the formula holds for $k + 1$, i.e., $P(k+1) = (k+1) + 1 = k + 2$

Using the inductive hypothesis: $P(k+1) = P(k) \times \left(1 + \frac{1}{k+1}\right)$ Substitute the inductive hypothesis $P(k) = k + 1$: $P(k+1) = (k + 1) \times \left(1 + \frac{1}{k+1}\right) = (k + 1) \times \frac{k+2}{k+1} = k + 2$

Thus, the formula holds for $k + 1$, completing the inductive step.

By mathematical induction, the formula $P(n) = n + 1$ is true for all $n \geq 1$.

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Would you like more details or have any questions about this process?

Here are 5 relative questions for further exploration:

1. How does the product change if the terms were $1 - \frac{1}{n}$ instead of $1 + \frac{1}{n}$?
2. What happens if the sum instead of the product is taken for the same terms?
3. How can this result be generalized for non-integer values of $n$?
4. Can this formula be extended to complex numbers?
5. What role does the harmonic series play in this type of product?

Tip: Mathematical induction is a powerful technique, often used to prove statements for sequences or recursive relations. Familiarize yourself with its base case and inductive step structure.